Arrange the following carbanions in order of their decreasing stability.

A. ${\mathrm{H}}_{3}\mathrm{C}-\mathrm{C}\equiv {\mathrm{C}}^{-}$

B. $\mathrm{H}-\mathrm{C}\equiv {\mathrm{C}}^{-}$

C. ${\mathrm{H}}_{3}\mathrm{C}-\mathrm{C}{\stackrel{-}{\mathrm{H}}}_{2}$

1. A>B>C

2. B>A>C

3. C>B>A

4. C>A>B

Hint: As the percentage of s character increases electronegativity of carbon increase

EXPLANATION

+I effect decreases the stability of carbon anion. Since CH3- group has + I-effect, therefore, it intensifies the negative charge and hence destabilises (A) relative to (B). sp hybridised carbanion is more stabilised than sp3 hybridized carbanion.

Thus, the stability order is as follows:
$\underset{{\left(\mathrm{B}\right)}^{\mathrm{sp}}}{\mathrm{CH}\equiv {\mathrm{C}}^{-}}>\underset{{\left(\mathrm{A}\right)}^{\mathrm{sp}}}{{\mathrm{CH}}_{3}-\mathrm{C}\equiv {\mathrm{C}}^{-}}>\underset{{\left(\mathrm{C}\right)}^{{\mathrm{sp}}^{3}}}{{\mathrm{CH}}_{3}-\stackrel{-}{\mathrm{C}}{\mathrm{H}}_{2}}$

Hence, B>A>C